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Is a mixed second rank tensor reducible?

  1. Jul 21, 2009 #1
    As a complete novice, I'm reading a text which says that a mixed second rank tensor [tex]T^{u}_{v}[/tex] is reducible but don't see how. Anyone care to show me? :wink:
     
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  3. Jul 21, 2009 #2

    robphy

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    What is the definition of reducible? (I'm not sure what you mean.)
     
  4. Jul 21, 2009 #3
    It means it can be broken down into parts that transform among themselves.
     
  5. Jul 21, 2009 #4

    robphy

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    Can you provide the reference where the statement in your first post appears?
     
  6. Jul 22, 2009 #5
    It comes from Anderson's Principles of Relativty Physics on page 19:

    " Thus from the components of a tensor T[tex]^{uv}[/tex] we can construct its symmetric part T[tex]^{(uv)}[/tex] and its antisymmetric part T[tex]^{[uv]}[/tex] according to

    [tex]T^{(uv)} = 1/2(T^{uv}+ T^{vu})[/tex]

    [tex]T^{[uv]} = 1/2(T^{uv}- T^{vu})[/tex]

    Similarly we can construct the transformed symmetric part T[tex]^{'(uv)}[/tex] and the antisymmetric part T[tex]^{'[uv]}[/tex] from the transformed T[tex]^{'uv}[/tex]. Then, one can show that T[tex]^{'(uv)}[/tex] is a function of T[tex]^{(uv)}[/tex] and the mapping function only, and similarly for T[tex]^{[uv]}[/tex]. When ever a geometrical object can be broken up into parts that transform among themselves, we say that we have a reducible object. If no such decomposition as possible. we have an irreducible object.
     
  7. Jul 22, 2009 #6

    robphy

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    OK, I see now.
    I take it you are referring to Problem 1.7....
    where they are telling you to "decompose" [itex]T_\nu{}^\mu[/itex]
    into irreducible parts consisting of
    its "trace" [itex]T_\mu{}^\mu[/itex] and
    a "traceless " (or "trace-free") tensor (which they give as [itex]T_\nu{}^\mu-\frac{1}{4}\delta_\nu{}^\mu T_\rho{}^\rho[/itex] on the assumption one is working in a 4-dimensional space).

    So, basically they have told you to write [itex]T_\nu{}^\mu[/itex]
    as a sum of two tensors, one of which they gave you:
    [itex]T_\nu{}^\mu = \mbox{(trace-part tensor)}_\nu{}^\mu + \mbox{(trace-free tensor)}_\nu{}^\mu [/itex]
     
  8. Jul 22, 2009 #7
    Yes. The question says: "show that it is reducible *and* its irreducuble parts are...". So I would have thought you should be able to first show its reducible from the transformation properties of a mixed second rank tensor, rather than being told what the irreducible components are and from that showing its reducible :smile
     
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